Free fall ellipse or parabola?
$begingroup$
Herbert Spencer somewhere says that the parabola of a ballistic object is actually a portion of an ellipse that is indistinguishable from a parabola--is that true? It would seem plausible since satellite orbits are ellipses and artillery trajectories are interrupted orbits.
newtonian-mechanics newtonian-gravity orbital-motion projectile free-fall
edited Apr 1 at 4:49
Qmechanic♦
107k121991239
asked Mar 31 at 23:28
user56930user56930
15726
$endgroup$
add a comment |
$begingroup$
Herbert Spencer somewhere says that the parabola of a ballistic object is actually a portion of an ellipse that is indistinguishable from a parabola--is that true? It would seem plausible since satellite orbits are ellipses and artillery trajectories are interrupted orbits.
newtonian-mechanics newtonian-gravity orbital-motion projectile free-fall
edited Apr 1 at 4:49
Qmechanic♦
107k121991239
asked Mar 31 at 23:28
user56930user56930
15726
$endgroup$
add a comment |
$begingroup$
Herbert Spencer somewhere says that the parabola of a ballistic object is actually a portion of an ellipse that is indistinguishable from a parabola--is that true? It would seem plausible since satellite orbits are ellipses and artillery trajectories are interrupted orbits.
newtonian-mechanics newtonian-gravity orbital-motion projectile free-fall
edited Apr 1 at 4:49
Qmechanic♦
107k121991239
asked Mar 31 at 23:28
user56930user56930
15726
$endgroup$
Herbert Spencer somewhere says that the parabola of a ballistic object is actually a portion of an ellipse that is indistinguishable from a parabola--is that true? It would seem plausible since satellite orbits are ellipses and artillery trajectories are interrupted orbits.
newtonian-mechanics newtonian-gravity orbital-motion projectile free-fall
newtonian-mechanics newtonian-gravity orbital-motion projectile free-fall
edited Apr 1 at 4:49
Qmechanic♦
107k121991239
asked Mar 31 at 23:28
user56930user56930
15726
edited Apr 1 at 4:49
Qmechanic♦
107k121991239
asked Mar 31 at 23:28
user56930user56930
15726
edited Apr 1 at 4:49
Qmechanic♦
107k121991239
edited Apr 1 at 4:49
Qmechanic♦
107k121991239
edited Apr 1 at 4:49
Qmechanic♦
107k121991239
107k121991239
asked Mar 31 at 23:28
user56930user56930
15726
asked Mar 31 at 23:28
user56930user56930
15726
asked Mar 31 at 23:28
user56930user56930
15726
15726
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.
On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.
At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.
answered Mar 31 at 23:35
Cort AmmonCort Ammon
24.4k34881
$endgroup$
4
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
Apr 1 at 1:48
3
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
Apr 1 at 8:42
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
Apr 2 at 4:52
add a comment |
$begingroup$
One easy way to tell the difference between a highly eccentric elliptical orbit and a true parabolic orbit is that an object in a parabolic orbit travels at its escape velocity exactly. In astronomy, such orbits are as rare as circular orbits, i.e. they don't exist. An object well below the escape velocity can be in an elliptical orbit that has an eccentricity very close to 1, making it look much like a parabolic orbit when only part of the curve is examined.
A relatively slow projectile on the surface of the Earth is actually a closed curve ellipse, and if the Earth got out of its way by shrinking to the size of basketball with the same gravitational field, the object would return to its original place in a long cigar shaped elliptical path.
As an aside, if an object is traveling faster than its escape velocity, it is in a hyperbolic orbit.
answered Apr 1 at 7:46
Bill WattsBill Watts
39217
$endgroup$
2
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
Apr 1 at 16:31
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
Apr 1 at 19:47
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
Apr 2 at 12:11
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.
On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.
At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.
answered Mar 31 at 23:35
Cort AmmonCort Ammon
24.4k34881
$endgroup$
4
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
Apr 1 at 1:48
3
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
Apr 1 at 8:42
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
Apr 2 at 4:52
add a comment |
$begingroup$
The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.
On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.
At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.
answered Mar 31 at 23:35
Cort AmmonCort Ammon
24.4k34881
$endgroup$
4
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
Apr 1 at 1:48
3
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
Apr 1 at 8:42
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
Apr 2 at 4:52
add a comment |
$begingroup$
The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.
On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.
At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.
answered Mar 31 at 23:35
Cort AmmonCort Ammon
24.4k34881
$endgroup$
The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.
On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.
At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.
answered Mar 31 at 23:35
Cort AmmonCort Ammon
24.4k34881
answered Mar 31 at 23:35
Cort AmmonCort Ammon
24.4k34881
answered Mar 31 at 23:35
Cort AmmonCort Ammon
24.4k34881
answered Mar 31 at 23:35
Cort AmmonCort Ammon
24.4k34881
24.4k34881
4
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
Apr 1 at 1:48
3
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
Apr 1 at 8:42
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
Apr 2 at 4:52
add a comment |
4
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
Apr 1 at 1:48
3
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
Apr 1 at 8:42
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
Apr 2 at 4:52
4
4
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
Apr 1 at 1:48
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
Apr 1 at 1:48
3
3
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
Apr 1 at 8:42
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
Apr 1 at 8:42
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
Apr 2 at 4:52
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
Apr 2 at 4:52
add a comment |
$begingroup$
One easy way to tell the difference between a highly eccentric elliptical orbit and a true parabolic orbit is that an object in a parabolic orbit travels at its escape velocity exactly. In astronomy, such orbits are as rare as circular orbits, i.e. they don't exist. An object well below the escape velocity can be in an elliptical orbit that has an eccentricity very close to 1, making it look much like a parabolic orbit when only part of the curve is examined.
A relatively slow projectile on the surface of the Earth is actually a closed curve ellipse, and if the Earth got out of its way by shrinking to the size of basketball with the same gravitational field, the object would return to its original place in a long cigar shaped elliptical path.
As an aside, if an object is traveling faster than its escape velocity, it is in a hyperbolic orbit.
answered Apr 1 at 7:46
Bill WattsBill Watts
39217
$endgroup$
2
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
Apr 1 at 16:31
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
Apr 1 at 19:47
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
Apr 2 at 12:11
add a comment |
$begingroup$
One easy way to tell the difference between a highly eccentric elliptical orbit and a true parabolic orbit is that an object in a parabolic orbit travels at its escape velocity exactly. In astronomy, such orbits are as rare as circular orbits, i.e. they don't exist. An object well below the escape velocity can be in an elliptical orbit that has an eccentricity very close to 1, making it look much like a parabolic orbit when only part of the curve is examined.
A relatively slow projectile on the surface of the Earth is actually a closed curve ellipse, and if the Earth got out of its way by shrinking to the size of basketball with the same gravitational field, the object would return to its original place in a long cigar shaped elliptical path.
As an aside, if an object is traveling faster than its escape velocity, it is in a hyperbolic orbit.
answered Apr 1 at 7:46
Bill WattsBill Watts
39217
$endgroup$
2
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
Apr 1 at 16:31
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
Apr 1 at 19:47
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
Apr 2 at 12:11
add a comment |
$begingroup$
One easy way to tell the difference between a highly eccentric elliptical orbit and a true parabolic orbit is that an object in a parabolic orbit travels at its escape velocity exactly. In astronomy, such orbits are as rare as circular orbits, i.e. they don't exist. An object well below the escape velocity can be in an elliptical orbit that has an eccentricity very close to 1, making it look much like a parabolic orbit when only part of the curve is examined.
A relatively slow projectile on the surface of the Earth is actually a closed curve ellipse, and if the Earth got out of its way by shrinking to the size of basketball with the same gravitational field, the object would return to its original place in a long cigar shaped elliptical path.
As an aside, if an object is traveling faster than its escape velocity, it is in a hyperbolic orbit.
answered Apr 1 at 7:46
Bill WattsBill Watts
39217
$endgroup$
One easy way to tell the difference between a highly eccentric elliptical orbit and a true parabolic orbit is that an object in a parabolic orbit travels at its escape velocity exactly. In astronomy, such orbits are as rare as circular orbits, i.e. they don't exist. An object well below the escape velocity can be in an elliptical orbit that has an eccentricity very close to 1, making it look much like a parabolic orbit when only part of the curve is examined.
A relatively slow projectile on the surface of the Earth is actually a closed curve ellipse, and if the Earth got out of its way by shrinking to the size of basketball with the same gravitational field, the object would return to its original place in a long cigar shaped elliptical path.
As an aside, if an object is traveling faster than its escape velocity, it is in a hyperbolic orbit.
answered Apr 1 at 7:46
Bill WattsBill Watts
39217
answered Apr 1 at 7:46
Bill WattsBill Watts
39217
answered Apr 1 at 7:46
Bill WattsBill Watts
39217
answered Apr 1 at 7:46
Bill WattsBill Watts
39217
39217
2
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
Apr 1 at 16:31
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
Apr 1 at 19:47
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
Apr 2 at 12:11
add a comment |
2
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
Apr 1 at 16:31
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
Apr 1 at 19:47
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
Apr 2 at 12:11
2
2
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
Apr 1 at 16:31
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
Apr 1 at 16:31
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
Apr 1 at 19:47
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
Apr 1 at 19:47
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
Apr 2 at 12:11
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
Apr 2 at 12:11
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
